This is super belated, but I'll try to give an answer to this question, should it be helpful for future homework doers. As is, the question is a bit under-specified because it may not be evident what frequency is being mentioned (resonance, precession, etc., though essentially the same number), which is probably why answers have not been forthcoming.
The short answer is YES, your approach is correct, but I'll make a reasonable assumption of precessional motion being detected, and explain why, based on the assumptions. In this experiment, one uses the Helmholz coil to do the cancellation of Earth's field (plus any other sources), but the magnet is used to provide detection. Basically, a magnet at some small angle with a total net magnetic field $B_T$ will undergo what is known as 'precession' (wobble) about the total magnetic field at a frequency $\omega_p$ proportional to
$\omega_p = 2\pi f \propto \gamma * B_T$
$\gamma$ is called the gyromagnetic ratio and is proportional to charge over mass. $B_T$ is the net field including the Earth's magnetic field, any stray field sources from magnets, and the field produced by the coil current. Your equation above is a form of $(1/2\pi)\gamma * B_T$, and/or the corresponding classical energy, proportional to $\omega^2$. So, when you cancel the earths magnetic field (plus any other stray magnetic field source), it will be observed when the precessional frequency vanishes or becomes zero, or the magnet's energy vanishes (ceases motion). Under these assumptions, your approach is correct :) Hopefully this explains a bit more as to why. So, well done, even though you are probably a Professor by now!
The argument below is paraphrased from Zangwill, Section 10.2.2. Set the origin of our coordinates at the spot we're trying to find the field, and let the axis of the solenoid be the $z$-axis. Consider a horizontal "slice" of the solenoid of width dz at a height $z$ above the xy-plane. By the Biot-Savart Law, the magnetic field due to this slice is just that of a current loop:
$$
d\vec{B} = - \frac{\mu_0 K dz}{4 \pi} \oint \frac{d\vec{\ell} \times \hat{r}}{r^2} = - \frac{\mu_0 K dz}{4 \pi} \oint \frac{d\vec{\ell} \times \vec{r}}{r^3}.
$$
(The minus sign is there because $\vec{r}$ in this formula denotes the source point, not the field point.) The slice can be thought of as a parametric curve $\vec{\ell}(\lambda)$, and it can be related to $\vec{r}$ by $\vec{r} = \vec{\ell} + z \hat{z}$, where $\vec{\ell}$ is in the xy-plane. Since $\vec{\ell}$ is perpendicular to $\hat{z}$ by construction, we have $r = \sqrt{\ell^2 + z^2}$, and so we can write the full integral as
\begin{equation}
d\vec{B} = - \frac{\mu_0 K dz}{4 \pi} \oint \frac{d\vec{\ell} \times (\vec{\ell} + z \hat{z})}{(\ell^2 + z^2)^{3/2}} = - \frac{\mu_0 K dz}{4 \pi} \left[ \oint \frac{d\vec{\ell} \times \vec{\ell} }{(\ell^2 + z^2)^{3/2}} + z \oint \frac{d\vec{\ell} \times \hat{z} }{(\ell^2 + z^2)^{3/2}} \right ]
\end{equation}
When we integrate from $z = -\infty$ to $\infty$, the second term will vanish (because it is odd with respect to $z$, while the first term evaluates to $2/\ell^2$. Thus, we now have
$$
\vec{B} = -\frac{\mu_0 K}{2 \pi} \oint \frac{d\vec{\ell} \times \vec{\ell} }{\ell^2}.
$$
Writing $\vec{\ell}$ in cylindrical coordinates, we have $\vec{\ell} = s \hat{s}$ and $d\vec{\ell} = ds \hat{s} + s \, d\phi \hat{\phi}$; thus, $-d\vec{\ell} \times \vec{\ell} = -s^2 \, d\phi \hat{z}$, and we have
$$
\vec{B} = \frac{\mu_0 K}{2 \pi} \hat{z} \oint \, d\phi.
$$
The integral is now the net solid angle traversed when we go around the loop. If the curve $\vec{\ell}(\lambda)$ encloses the origin (i.e., we are inside the solenoid), this will be $2\pi$; otherwise, it will be zero. Thus,
$$
\vec{B} = \begin{cases} \mu_0 K \hat{z} & \text{inside} \\
0 & \text{outside}.\end{cases}
$$
Note that this derivation did not assume any particular shape for the cross-section of the solenoid (i.e., the shape of the curve $\vec{\ell}$.)
To preemptively address any concerns about infinite solenoids being unrealistic: how would this derivation be modified if we had a non-infinite (but long) solenoid? Return to the step where we had written $d\vec{B}$ in terms of $dz$ and two integrals. If we assume that the solenoid now stretches from $z_1$ to $z_2$ instead, the expression becomes:
$$
\vec{B} = - \frac{\mu_0 K}{4 \pi} \left[ \oint \frac{\sigma(z) \, d\vec{\ell} \times \vec{\ell} }{\ell^2 \sqrt{(\ell/z)^2 + 1}} - \oint \frac{d\vec{\ell} \times \hat{z} }{|z| \sqrt{(\ell/z)^2 + 1}} \right ]_{z = z_1}^{z_2}.
$$
where $\sigma(z)$ is the sign of $z$. While we can't evaluate these integrals in as much generality, we still retain some important features. The first term will still point in the z-direction regardless of $z_1$ and $z_2$, and the second term will always point in the xy-plane. Moreover, the second term still vanishes so long as $z_1 = - z_2$. Thus, at the midpoint of a long solenoid, the external field still points along the axis of the solenoid (as it must by symmetry.)
If we wanted, we could expand these integrals in a power series in $\ell/z_1$ and $\ell/z_2$, assuming this quantity is small; this is saying that the perpendicular distance from the field point to the solenoid is much smaller than the distance to its ends. In this limit, we would recover the infinite-solenoid result plus corrections due to the finite length of the solenoid. The leading-order correction to the horizontal field would be $\mathcal{O}(\ell/z)$, while the leading-order correction to the vertical field will be $\mathcal{O}(\ell/z)^2$.
Best Answer
I tried to address this question in a short article that recently got published in the European Journal of Physics. https://arxiv.org/abs/1610.07876
You can also look at this article by Farley, Price, which uses a nice argument and physically motivates the problem (American Journal of Physics 69, 751 (2001); doi 10.1119/1.1362694): https://doi.org/10.1119/1.1362694
Also, there is an exact calculation done in this old NASA Technical Note D-465: http://paginas.fe.up.pt/~ee08173/wp-content/uploads/2014/03/finite-solenoid.pdf